Let A = [ \beginmatrix i & - i - i & i \endmatrix ],i = \sqrt - 1 . Then, the system of linear equations A^8[ \beginmatrix x y \endmatrix ] = [ \beginmatrix 8 64 \endmatrix ] has :

Matrices and Determinants PYQ — JEE Main 2021 · 16 Mar · Shift 1 · Q28 with Solution

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Correct answer: D

$A=\begin{bmatrix} i & -i \\ -i & i \end{bmatrix}=i\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$

Let $B=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$ so that $A=iB.$

First,

=\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}=2B.$$ Hence, $$B^2=2B.$$ This implies by induction that for $n\ge 1$, $$B^n=2^{n-1}B.$$ So, $$B^8=2^7B=128B.$$ --- 3. **Compute $A^8$** Since $A=iB$, $$A^8=(iB)^8=i^8B^8.$$ Now, $$i^8=(i^4)^2=1.$$ Therefore, $$A^8=B^8=128\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} =\begin{bmatrix} 128 & -128 \\ -128 & 128 \end{bmatrix}.$$ --- 4. **Form the system** We need to solve $$A^8\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8\\64\end{bmatrix}.$$ That is, $$\begin{bmatrix} 128 & -128 \\ -128 & 128 \end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8\\64\end{bmatrix}.$$ This gives the equations: $$128x-128y=8$$ $$-128x+128y=64$$ Dividing by $128$: $$x-y=\frac{1}{16}$$ $$-x+y=\frac{1}{2}$$ But the second equation is equivalent to $$x-y=-\frac{1}{2},$$ which contradicts $$x-y=\frac{1}{16}.$$ So the system is **inconsistent**. --- 5. **Conclusion** Hence, the system has **no solution**. Therefore, the correct option is: $$\boxed{\text{D: No solution}}$$ --- 6. **Comparison with stored answer** Stored correct answer: **D** My derived answer: **D** They agree.

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Correct answer: D

$A=\begin{bmatrix} i & -i \\ -i & i \end{bmatrix}=i\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$

Let $B=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$ so that $A=iB.$

First,

=\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}=2B.$$ Hence, $$B^2=2B.$$ This implies by induction that for $n\ge 1$, $$B^n=2^{n-1}B.$$ So, $$B^8=2^7B=128B.$$ --- 3. **Compute $A^8$** Since $A=iB$, $$A^8=(iB)^8=i^8B^8.$$ Now, $$i^8=(i^4)^2=1.$$ Therefore, $$A^8=B^8=128\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} =\begin{bmatrix} 128 & -128 \\ -128 & 128 \end{bmatrix}.$$ --- 4. **Form the system** We need to solve $$A^8\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8\\64\end{bmatrix}.$$ That is, $$\begin{bmatrix} 128 & -128 \\ -128 & 128 \end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8\\64\end{bmatrix}.$$ This gives the equations: $$128x-128y=8$$ $$-128x+128y=64$$ Dividing by $128$: $$x-y=\frac{1}{16}$$ $$-x+y=\frac{1}{2}$$ But the second equation is equivalent to $$x-y=-\frac{1}{2},$$ which contradicts $$x-y=\frac{1}{16}.$$ So the system is **inconsistent**. --- 5. **Conclusion** Hence, the system has **no solution**. Therefore, the correct option is: $$\boxed{\text{D: No solution}}$$ --- 6. **Comparison with stored answer** Stored correct answer: **D** My derived answer: **D** They agree.